# How do you find the intervals of increasing and decreasing given #y=x^2/(4x+4)#?

##### 1 Answer

Please see below.

#### Explanation:

The derivative of a function

- If
#y'>0# , then#y# is increasing. - If
#y'<0# , then#y# is decreasing.

So, we need to find the derivative of this function. We will need to use the quotient rule, which states that the derivative of the quotient of the functions

#y=u/v" "=>" "y'=((u'*v)-(u*v'))/v^2#

So, for the given function, we see that:

#y'=((d/dxx^2)(4x+4)-(x^2)(d/dx(4x+4)))/(4x+4)^2#

Through the power rule,

#y'=((2x)(4x+4)-x^2(4))/(4x+4)^2=(4x^2+8x)/(4x+4)^2#

Taking

#y'=(4x(x+2))/(16(x+1)^2)=(x(x+2))/(4(x+1)^2)#

So, we want to find when this is positive and when this is negative. Approach this by first finding the spots where the derivative could *switch* from positive to negative. That is, when the derivative equals

The derivative equals

So, we want to examine the sign of the derivative *around* the three points

When

For

For

For

#color(red)star# We can now conclude that#y# is decreasing on#(-2,-1)uu(-1,0)# and is increasing on#(-oo,-2)uu(0,oo)# .

Note that the decreasing intervals **cannot** be combined into

graph{x^2/(4x+4) [-10, 10, -5, 5]}